A series with same common difference is known as arithmetic series. The first term of series is ‘a‘ and common difference is d. The series looks like a, a + d, a + 2d, a + 3d, . . . Find the sum of series.具有相同共同差異的系列被稱為算術系列。系列的第一個術語是“ a ”，共同的區別是d。該系列看起來像a + d，a + 2d，a + 3d。。。找到系列的總和。

Input : a = 1
        d 
 
= 2
        n = 4
Output : 16
1 + 3 + 5 + 7 = 16

Input : a = 2.5
        d = 1.5
        n = 20
Output : 335

Input:
The first line consists of an integer T i.e number of test cases. The first line and only line of each test case consists of three integers a,d,n.

輸入：
第一行由整數T即測試用例數組成。每個測試用例的第一行和第一行由三個整數

Output:
Print the sum of the series. With two decimal places.

輸出：
打印系列的總和。有兩位小數。

Constraints:
1<=T<=100
1<=a,d,n<=1000

約束：
1 <= T <= 100
1 <= a，d，n <= 1000

Example:

Input:
2
1 2 4
2.5 1.5 20

Output:
16.00
335.00

其實，就是一個等差數列的求和問題，不過需要註意的是輸出是兩位小數。

下面是我的代碼實現：

#include <stdio.h>
#include 
 
<stdlib.h>

int main()
{
    int n,i,j;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        float a,d,n;
        scanf("%f %f %f",&a,&d,&n);
        printf("%.2f\n",n*a+n*(n-1)*d/2);
    }
    return 0;
}

然後竟然出現了錯誤？一起來看一下：

好的，發現了問題應該在類型轉換上面。

如有疑問，請留言。

Sum of AP series——AP系列之和